A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates.Now, an insulating slab of same area but thickness d / 2 is inserted between the plates as shown infigure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be

$C_{0} = \frac{ε_{0} A}{d}$ … (i)

According to second condition,

A' = A, t = d / 2, K = 4

$∴$ Capacitance, $C = \frac{ε_{0} A}{(d - t) + \frac{t}{K}} = \frac{ε_{0} A}{(d - \frac{d}{2}) + \frac{d / 2}{4}}$

$= \frac{ε_{0} A}{\frac{d}{2} + \frac{d}{8}} = \frac{8}{5} \cdot \frac{ε_{0} A}{d}$

$∴$ $\frac{C}{C_{0}} = \frac{\frac{8}{5} \cdot \frac{ε_{0} A}{d}}{\frac{ε_{0} A}{d}}$

$\Rightarrow \frac{C}{C_{0}} = \frac{8}{5}$

$\Rightarrow C : C_{0} = 8 : 5$

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